Is the surface of a heavy sphere bigger than $4 \pi r^2$ due to general relativity? I am unfortunately not familiar with the mathematics behind general relativity. However, on a heavy planet (say a sphere) gravity will bend space-time in a way that an object initially in rest, will experience over time, more and more of its time-dimension as a space-dimension towards the planet, hence causing it to accelerate towards the planet. (Video: https://www.youtube.com/watch?v=jlTVIMOix3I )
If I generalize this line of thought conceptually, this would mean that on the surface of the planet, there is more space (and less time) than an outside observer assuming a flat space-time would expect.
Is this correct? Is the surface of a heavy spherical planet bigger than $4 \pi r^2$?
This does seem to match with the rubber sheet visualisations of curved space-time, which show a negative curved space around heavy objects. It also seems to match with the idea that time slows down falling in a black hole, while space tends to go to infinity.
But is this line of thought correct?
Because I'm also inclined to think that the all time-dimension are becoming space in the direction towards the planet. Therefore, there is more space around heavy objects, but not in a direction perpendicular to the direction of gravity. Therefore, the surface of a spherical planet would still be exactly $4 \pi r^2$, since all normal vectors on this surface are parallel to the direction of gravity.
So, if I would pull a rope through the planet and measure its length $2r$, and measure the surface of the planet $A$ walking around on the planet, $A>4\pi r^2$. Yes?
 A: Yes. The radial distance you measure does not jive as you move outwards. In other words if you move out by 100km the surface area will not be 4π(R + 100km)2, but a little less. 
There are two 'r' here. One is the distance covered on the trip from the centre to the surface (say 'r'), the other is determined by measuring the area of the planet and using the 4πR2 rule. These will not be equal as they would in flat space. r will be bigger than R. 
Or is this exactly backwards from what you are thinking? I guess it depends on which r you are using in your question.
A: Short answer is yes (at least according to GR): despite the fact that your reasoning is not rigorous, you came to the right conclusion.
It can be seen mathematically. Take one of the idealized solutions of GR (the most simple one is the Schwarzschild metric) and evaluate the integral over the sphere with radial coordinate $r$. Then integrate the distance from $0$ to $r$ - the radius of the sphere (note that it is not equal to $r$ since the radial coordinate is not physical - it is just a quantity we introduce). Compare these results and it will lead you to the same conclusion.
A: A little over a year later, I think I can answer my own question.
If you would measure the radius of a heavy object by pulling a cord through it, and compare it to the surface or the circumference, the values would not fit, and $A<4\pi r^2$ or $d<2\pi r$, because space is stretched in the radial direction.
However, if you would measure the circumference $C$ and the area $A$ of the sphere, then $A \pi=C^2$, exactly like you would expect in flat space, as space is only stretched in the radial direction, normal to the surface of the heavy object.
This was the question I like to have answered back then, but I formulated it rather clumsily.
